Question Number 100330 by bemath last updated on 26/Jun/20

Commented bybobhans last updated on 26/Jun/20

(1)+(2)+(3) ⇒x^2 +y^2 +z^2 +2(xy+xz+yz)=16  (x+y+z)^2 =16 ⇒ { ((x+y+z=4)),((x+y+z=−4)) :}  case:1   •x(x+y+z) = 1 ⇒x=(1/4)  •y(x+y+z)= 6 ⇒y=(3/2)  •z(x+y+z) = 9 ⇒ z=(9/4)